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May 19, 2010 at 20:53 comment added Torsten Ekedahl I understand, I was thinking about the possibility that you meant zero-cycles but was thrown off by the fact that rational equivalence doesn't work. If I remember correctly the equivalence relation you are talking about is "abelian equivalence", the equivalence relation generated by maps into torsors over abelian varieties. This gives the quite tautological relation with the definition I was referring to. Of course there is a formal definition of the degree $1$ part, in that a torsor over $A$ gives an extension $0\to A \to A' \to \mathbb Z \to 0$ with the torsor the inverse image of $1$.
May 19, 2010 at 20:07 vote accept Simon Rose
May 19, 2010 at 20:04 comment added Pete L. Clark For some more details, see Sections 4.1 and 4.3 of math.uga.edu/~pete/wc2.pdf. (Nevertheless these concepts are not due to me.)
May 19, 2010 at 20:02 comment added Pete L. Clark For a smooth projective variety $V$, one has a total Albanese scheme $\operatorname{Alb}(V)$, whose points parameterize all zero-cycles on $V$. This comes with a degree map to $\mathbb{Z}$, and $\operatorname{Alb}^i(V)$ is the component consisting of zero-cycles of degree $i$. Each $\operatorname{Alb}^i(V)$ is -- in an evident way -- a torsor under $\operatorname{Alb}^0(V)$.
May 19, 2010 at 19:53 comment added Torsten Ekedahl What do you mean by "degree 1 Albanese variety"? The definition I know of the Albanese variety of a (smooth or normal say) proper variety $X$ is as a universal map $X \to V$ to torsors over abelian varieties. Hence, in your case the Albanese map is the identity map (which is what you say) but I do not understand what you mean by degree $1$ and degree $0$.
May 19, 2010 at 18:42 history answered Pete L. Clark CC BY-SA 2.5